Hint: There are two types of $4$ digit odd integers with all digits different: (i) First digit is even ($2,4,6,8$) or (ii) first digit is odd. Count the Type (i) numbers, the Type (ii) numbers, and add.
Type (i): There are $4$ choices for the first digit. For each of these choices there are $5$ choices for the last digit. For each of the choices we have made so far, there are $8$ choices for the second digit, and for each such choice $7$ choices for the third, a total of $(4)(5)(8)(7)$.
Now it's your turn: make a similar calculation for Type (ii).
The analysis for the even numbers will be roughly similar. The two cases are (i) last digit is $0$ and (ii) last digit is not $0$.
Remark: The tricky thing is that the first digit cannot be $0$. A slightly nicer approach, I think, is to first allow $0$ as a first digit, and then take away the "numbers" with first digit $0$, which we should not have counted.
So for odd numbers, if we allow $0$ as a first digit, we can choose the last digit in $5$ ways, and then for the rest we have $(9)(8)(7)$ choices, for a total of $(5)(9)(8)(7)$.
Now if we have $0$ as the first digit, the rest can be filled in $(5)(8)(7)$ ways.
So our answer is $(5)(9)(8)(7)-(5)(8)(7)=2240$.
For the evens, the same analysis gives $(5)(9)(8)(7)-(4)(8)(7)$.
Best Answer
Usual tricks to get started for any problem
Have you any ideas on how to do any of these three things?
For counting problems in particular, one trick that is:
If you are having any troubles at all, it's nearly always worth at least writing down the complementary problem to see if it offers any inspiration.